Question

E (t) = Eo cos (kz-wt) Where k is the wave-number and w is the angular frequency of the wave. Now some scientist tells you that a distant star has been observed to be green in color, with an observed frequency of fobserved 550 THz (terahertz, 1012). She also says that from other factors, it is known that the star should be emitting light with wavelength factual-1800 THz somewhere in the UV spectrum. They ask you to find two things What is the equation that describes the electromagnetic wave? Fill in as many parameters as vou can. ii) How fast is this star moving relative to the Earth? iii) Estimate the power radiated by this star. The observatory's signal is about 2 μW m-2 and we know that the signal we received is exactly 100 years old. (First, find out how far away the star is!)

Answer 1

a) Actual frequency of source

a) Actual frequency of source f_actual = 1800 THz = 1800 * 10^12 Hz Wavelength of light emitted from the source is lambda = c/f_actual = 3 * 10^8/1800 * 10^12 = 1.667 * 10^-7 m Wave number k = 2 pi/1.667 * 10^-7 = 3.77 * 10^7 m^-1 Wave frequency omega = 2 pi f_actual = 2 pi * 1800 * 10^12 = 1.13 * 10^16 rad/s Wave equation is E^vector (t) = E_0^vector cos (3.77 * 10^7 x - 1.13 * 10^16 t) b) f_observed/f_actual = squareroot 1 - beta/1 + beta (f_observed/f_actual)^2 = 1 - beta/1 + beta (f_observed/f_actual)^2 (1 + beta) = 1 - beta (550/1800)^2 (1 + beta) = 1 - beta beta = 0.8292 v/c = 0.8292 Speed of star relative to earth is v = 0.8292c = 2.45 * 10^8 m/s c) Time taken for the light reach earth from the star is t = 100 years = 100 * 365 * 24 * 3600 s

Wavelength of light emitted from the source is

Wave number $k=\frac{2\pi}{1.667*10^{-7}}=3.77*10^7\,m^{-1}$

Wave frequency $\omega=2\pi f_{actual}=2\pi *1800*10^{12}=1.13*10^{16}\,rad/s$

Wave equation is

----------------

b)

$\frac{f_{observed}}{f_{actual}}=\sqrt{\frac{1-\beta}{1+\beta}}$

$\left ( \frac{f_{observed}}{f_{actual}} \right )^2={\frac{1-\beta}{1+\beta}}$

$\left ( \frac{f_{observed}}{f_{actual}} \right )^2(1+\beta)={{1-\beta}}$

$\left ( \frac{550}{1800} \right )^2(1+\beta)={{1-\beta}}$

$\beta=0.8292$

$\frac{v}{c}=0.8292$

Speed of star relative to earth is $v=0.8292\,c=2.45*10^8\,m/s$

------------------

c)

Time taken for the light reach earth from the star is $t=100\,years=100*365*24*3600\,s$

Distance of star from the earth is $r=ct=(3*10^8)(100*365*24*3600)=9.4607*10^{17}\,m$

Intensity of light $I=2\mu W/m^2=(2*10^{-6})\,W/m^2$

Power radiated by star is

---------------